Talk:PlanetPhysics/Example of Vector Potential

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: example of vector potential %%% Primary Category Code: 02.30.-f %%% Filename: ExampleOfVectorPotential.tex %%% Version: 2 %%% Owner: pahio %%% Author(s): pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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If the \htmladdnormallink{solenoidal}{http://planetphysics.us/encyclopedia/SolenoidalVectorField.html} \htmladdnormallink{vector}{http://planetphysics.us/encyclopedia/Vectors.html} \,$\vec{U} = \vec{U}(x,\,y,\,z)$\, is a homogeneous \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} of degree $\lambda$ ($\neq -2$),\, then it has the \htmladdnormallink{vector potential}{http://planetphysics.us/encyclopedia/SolenoidalVectorField.html} \begin{align} \vec{A} = \frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}, \end{align} where\, $\vec{r} = x\vec{i}\!+\!y\vec{j}\!+\!z\vec{k}$\, is the \htmladdnormallink{position vector}{http://planetphysics.us/encyclopedia/PositionVector.html}.

{\em Proof.}\, Using the entry nabla acting on products, we first may write $$\nabla\times(\frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}) = \frac{1}{\lambda\!+\!2}[(\vec{r}\cdot\nabla)\vec{U} -(\vec{U}\cdot\nabla)\vec{r}-(\nabla\cdot\vec{U})\vec{r} +(\nabla\cdot\vec{r})\vec{U}].$$ In the brackets the first product is, according to Euler's \htmladdnormallink{theorem}{http://planetphysics.us/encyclopedia/Formula.html} on homogeneous functions, equal to $\lambda\vec{U}$.\, The second product can be written as\, $U_x\frac{\partial\vec{r}}{\partial x}+ U_y\frac{\partial\vec{r}}{\partial y}+U_z\frac{\partial\vec{r}}{\partial z}$, which is $U_x\vec{i}+U_y\vec{j}+U_z\vec{k}$, i.e. $\vec{U}$.\, The third product is, due to the sodenoidalness, equal to\, $0\vec{r} = \vec{0}$.\, The last product equals to $3\vec{U}$ (see the \htmladdnormallink{first formula}{http://planetphysics.us/encyclopedia/PositionVector.html} for position vector).\, Thus we get the result $$\nabla\times(\frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}) = \frac{1}{\lambda\!+\!2}[\lambda\vec{U}-\vec{U}-\vec{0}+3\vec{U}] = \vec{U}.$$ This means that $\vec{U}$ has the vector potential (1).

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